Concept:Heavier doping provides a higher density of uncompensated ions per unit volume near the junction, satisfying the built-in potential barrier across a narrower physical distance.
Formula:$$W = \sqrt{\frac{2 \epsilon_s V_0}{q} \left( \frac{1}{N_A} + \frac{1}{N_D} \right)} \propto \frac{1}{\sqrt{N_{\text{doping}}}}$$
Solution:- As doping levels \( N_A \) and \( N_D \) increase, the required space-charge is established within a very narrow region close to the metallurgical interface.
- Therefore, heavier doping results in a thinner (narrower) depletion layer.
Why other options are incorrect:- Option B: Depletion width is inversely related to doping concentration; it narrows rather than expands.
- Option C: Depletion width depends strongly on doping density.
- Option D: The width decreases monotonically with increasing dopant concentration.
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